For a general total Hamiltonian, the bath and system-bath coupling don't respect the related symmetry of \(\hat{H}_S\) , in which case \(\hat{\rho }_G(t)\)  evolves into a mixed state where the decoherence happens. Nevertheless, what we concern here is whether the complete information of initial state, or at least the coherence, can be maintained when both the operators {\(\hat{a}_\alpha \) } and {\(\hat{\mathcal {O}}_{j}\) } respect the same symmetry as the system Hamiltonian \(\hat{H}_S\) .
As stated above, the fate of coherence is determine by unitarity or anti-unitarity of the symmetry. This conclusion had been pointed out in Ref.[1]}, [2]} and here we provide a brief explanation.
For the unitary symmetry, \(\hat{\Pi }_{G}\hat{\mathcal {O}}^\dagger \hat{\mathcal {O}}\hat{\Pi }_{G}\propto \hat{\Pi }_{G}\) , \(\hat{\Pi }_{G}\hat{\mathcal {O}}^\dagger \hat{\Pi }_{G}\propto \hat{\Pi }_{G}\) , and \(\hat{\Pi }_{G}\hat{\mathcal {O}}\hat{\Pi }_{G}\propto \hat{\Pi }_{G}\)  are guaranteed by Schur's lemma[3]} as long as \(\hat{\mathcal {O}}^\dagger \) , \(\hat{\mathcal {O}}\)  respect the same symmetry as \(\hat{H}_S\) . In this case, Lindblad superoperator \({L}\)  only acts on the subspace density matrix as \(\hat{\Pi }_{G}{L}[\hat{\rho }(t)]\hat{\Pi }_{G}\propto \rho _{G}(0)\) .
Accordingly, if we define
\(\rho _{G}(t)=\left(\begin{array}{cc}\rho _{++} & \rho _{+-}\\\rho _{-+} & \rho _{--}\end{array}\right)=\left(\begin{array}{cc}\langle \phi _{+}|\hat{\rho }|\phi _{+}\rangle & \langle \phi _{+}|\hat{\rho }|\phi _{-}\rangle \\\langle \phi _{-}|\hat{\rho }|\phi _{+}\rangle & \langle \phi _{-}|\hat{\rho }|\phi _{-}\rangle \end{array}\right),\) 
